Question

If 16 cot x = 12, then what will be the value of $\frac{\sin x-\cos x}{\sin x+\cos x}$

Answer 1

$\cot x=\frac{12}{16}\cot x=\frac{3}{4}$
Also, $\cot \theta =\frac{base}{perpendicular}$
So, we now construct a right triangle ABC, right angled at B such that $\angle BAC=\theta ,$ Base = 3 and perpendicular = 4
In $△ABC,$


$A{C}^2=A{B}^2+B{C}^{2}A{C}^2=(3{)}^2+(4{)}^{2}A{C}^2=9+16=25AC=5$
As we know $\sin \theta =\frac{perpendicular}{hypotenuse}\cos \theta =\frac{base}{hypotenuse}$
Therefore, $\sin x=\frac{BC}{AC}=\frac{4}{5}\cos x=\frac{AB}{AC}=\frac{3}{5}$
Now,
$\frac{\sin x-\cos x}{\sin x+\cos x}=\frac{\frac{4}{5}-\frac{3}{5}}{\frac{4}{5}+\frac{3}{5}}=\frac{\frac{4-3}{5}}{\frac{4+3}{5}}=\frac{1}{7}$